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can someone tell me how to do this? Find all solutions in the interval [0, 2π). 2 sin2x = sin x
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\[ 4 \sin(x) \cos(x)=\sin(x) \\ 4 \sin(x) \cos(x)-\sin(x)=0 \\ \sin(x) (4 \cos(x) -1)=0\\ \sin(x) =0 \\ 4 \cos(x) -1 =0\\ \cos(x) =\frac 1 4 \] Can you finish it now?
still confused >.<
When is sin(x) =0 for x between 0 and 2 Pi?
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